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2.3 Partielle Integration

Aus Online Mathematik Brückenkurs 2

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Zeile 25: Zeile 25:
To integrate products, one sometimes can make use of a method known as ''integration by parts''. The method is based on the reverse use of the rules for differentiation of products. If <math>f</math> and <math>g</math> are two differentiable functions then the rule for products gives
To integrate products, one sometimes can make use of a method known as ''integration by parts''. The method is based on the reverse use of the rules for differentiation of products. If <math>f</math> and <math>g</math> are two differentiable functions then the rule for products gives
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{{Fristående formel||<math>D\,(\,f\cdot g) = f^{\,\prime} \cdot g + f \cdot g'\,\mbox{.}</math>}}
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{{Displayed math||<math>D\,(\,f\cdot g) = f^{\,\prime} \cdot g + f \cdot g'\,\mbox{.}</math>}}
Now if one integrates both sides one gets
Now if one integrates both sides one gets
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{{Fristående formel||<math>f \cdot g = \int (\,f^{\,\prime} \cdot g + f \cdot g'\,)\,dx = \int f^{\,\prime} \cdot g\,dx + \int f\cdot g'\,dx</math>}}
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{{Displayed math||<math>f \cdot g = \int (\,f^{\,\prime} \cdot g + f \cdot g'\,)\,dx = \int f^{\,\prime} \cdot g\,dx + \int f\cdot g'\,dx</math>}}
or after re-ordering
or after re-ordering
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{{Fristående formel||<math>\int f^{\,\prime} \cdot g\,dx = f \cdot g - \int f \cdot g'\,dx\,\mbox{.}</math>}}
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{{Displayed math||<math>\int f^{\,\prime} \cdot g\,dx = f \cdot g - \int f \cdot g'\,dx\,\mbox{.}</math>}}
This gives us the formula for integration by parts.
This gives us the formula for integration by parts.
Zeile 39: Zeile 39:
<div class="regel">
<div class="regel">
'''Integration by parts:'''
'''Integration by parts:'''
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{{Fristående formel||<math>\int f(x)\cdot g(x)\,dx = F(x) \cdot g(x) - \int F(x) \cdot g'(x)\,dx\,\mbox{.}</math>}}
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{{Displayed math||<math>\int f(x)\cdot g(x)\,dx = F(x) \cdot g(x) - \int F(x) \cdot g'(x)\,dx\,\mbox{.}</math>}}
</div>
</div>
Zeile 55: Zeile 55:
If one chooses <math>f=x</math> and <math>g=\sin x</math> one gets <math>F=x^2/2</math> and <math>g'=\cos x</math>, and the formula for integration by parts gives
If one chooses <math>f=x</math> and <math>g=\sin x</math> one gets <math>F=x^2/2</math> and <math>g'=\cos x</math>, and the formula for integration by parts gives
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{{Fristående formel||<math>\int x \cdot \sin x \, dx = \frac{x^2}{2} \cdot \sin x - \int \frac{x^2}{2} \cdot \cos x \, dx\,\mbox{.}</math>}}
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{{Displayed math||<math>\int x \cdot \sin x \, dx = \frac{x^2}{2} \cdot \sin x - \int \frac{x^2}{2} \cdot \cos x \, dx\,\mbox{.}</math>}}
The new integral on the right-hand side in this case is not easier than the original integral.
The new integral on the right-hand side in this case is not easier than the original integral.
Zeile 61: Zeile 61:
If, instead, one chooses <math>f=\sin x</math> and <math>g=x</math> then <math>F=-\cos x</math> and <math>g'=1</math>, and
If, instead, one chooses <math>f=\sin x</math> and <math>g=x</math> then <math>F=-\cos x</math> and <math>g'=1</math>, and
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{{Fristående formel||<math>\int x \cdot \sin x \, dx = - x \cdot \cos x - \int - 1 \cdot \cos x \, dx = - x\cos x + \sin x + C\,\mbox{.}</math>}}
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{{Displayed math||<math>\int x \cdot \sin x \, dx = - x \cdot \cos x - \int - 1 \cdot \cos x \, dx = - x\cos x + \sin x + C\,\mbox{.}</math>}}
</div>
</div>
Zeile 72: Zeile 72:
Put <math>f=x^2</math> and <math>g=\ln x</math> since differentiation eliminates the logarithm when we carry out an integration by parts: <math>F=x^3/3</math> and <math>g'=1/x</math>. This gives us that
Put <math>f=x^2</math> and <math>g=\ln x</math> since differentiation eliminates the logarithm when we carry out an integration by parts: <math>F=x^3/3</math> and <math>g'=1/x</math>. This gives us that
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{{Fristående formel||<math>\begin{align*}\int x^2 \cdot \ln x \, dx &= \frac {x^3}{3} \cdot \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac {x^3}{3} \cdot \ln x - \frac{1}{3} \int x^2 \, dx\\[4pt] &= \frac{x^3}{3} \cdot \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C = \tfrac{1}{3}x^3 ( \ln x - \tfrac{1}{3} ) + C\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*}\int x^2 \cdot \ln x \, dx &= \frac {x^3}{3} \cdot \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac {x^3}{3} \cdot \ln x - \frac{1}{3} \int x^2 \, dx\\[4pt] &= \frac{x^3}{3} \cdot \ln x - \frac{1}{3} \cdot \frac{x^3}{3} + C = \tfrac{1}{3}x^3 ( \ln x - \tfrac{1}{3} ) + C\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Zeile 84: Zeile 84:
Put <math>f=e^x</math> and <math>g=x^2</math>, which gives that <math>F=e^x</math> and <math>g'=2x</math>, and an integration by parts gives
Put <math>f=e^x</math> and <math>g=x^2</math>, which gives that <math>F=e^x</math> and <math>g'=2x</math>, and an integration by parts gives
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{{Fristående formel||<math> \int x^2 e^x \, dx = x^2 e^x - \int 2x\,e^x \, dx\,\mbox{.}</math>}}
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{{Displayed math||<math> \int x^2 e^x \, dx = x^2 e^x - \int 2x\,e^x \, dx\,\mbox{.}</math>}}
This requires further integration by parts to solve the new integral <math>\,\int 2x\,e^x \, dx</math>. We choose in this case <math>f=e^x</math> and <math>g=2x</math>, which gives <math>F=e^x</math> and <math>g'=2</math>
This requires further integration by parts to solve the new integral <math>\,\int 2x\,e^x \, dx</math>. We choose in this case <math>f=e^x</math> and <math>g=2x</math>, which gives <math>F=e^x</math> and <math>g'=2</math>
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{{Fristående formel||<math>\int 2x\,e^x \, dx = 2x\,e^x - \int 2 e^x \, dx = 2x\,e^x - 2 e^x + C\,\mbox{.}</math>}}
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{{Displayed math||<math>\int 2x\,e^x \, dx = 2x\,e^x - \int 2 e^x \, dx = 2x\,e^x - 2 e^x + C\,\mbox{.}</math>}}
The original integral thus becomes
The original integral thus becomes
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{{Fristående formel||<math> \int x^2 e^x \, dx = x^2 e^x - 2x\,e^x + 2 e^x + C\,\mbox{.}</math>}}
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{{Displayed math||<math> \int x^2 e^x \, dx = x^2 e^x - 2x\,e^x + 2 e^x + C\,\mbox{.}</math>}}
</div>
</div>
Zeile 104: Zeile 104:
In a first integration by parts, we have chosen to integrate the factor <math>e^x</math> and differentiate the factor <math>\cos x</math>,
In a first integration by parts, we have chosen to integrate the factor <math>e^x</math> and differentiate the factor <math>\cos x</math>,
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{{Fristående formel||<math>\begin{align*}\int e^x \cos x \, dx &= e^x \cdot \cos x - \int e^x \cdot(-\sin x) \, dx\\[10pt] &= e^x \cos x + \int e^x \sin x \, dx\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*}\int e^x \cos x \, dx &= e^x \cdot \cos x - \int e^x \cdot(-\sin x) \, dx\\[10pt] &= e^x \cos x + \int e^x \sin x \, dx\,\mbox{.}\end{align*}</math>}}
The result of this is that we essentially have replaced the factor <math>\cos x</math> by <math>\sin x</math> in the integral. If we therefore use integration by parts once again (integrate the <math>e^x</math> and differentiate the <math>\sin x</math>) we get
The result of this is that we essentially have replaced the factor <math>\cos x</math> by <math>\sin x</math> in the integral. If we therefore use integration by parts once again (integrate the <math>e^x</math> and differentiate the <math>\sin x</math>) we get
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{{Fristående formel||<math>\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx\,\mbox{.}</math>}}
+
{{Displayed math||<math>\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx\,\mbox{.}</math>}}
Thus the original integral appears here again. Summarising we have:
Thus the original integral appears here again. Summarising we have:
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{{Fristående formel||<math>\int e^x \cos x \, dx = e^x \cos x + e^x \sin x - \int e^x \cos x \, dx</math>}}
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{{Displayed math||<math>\int e^x \cos x \, dx = e^x \cos x + e^x \sin x - \int e^x \cos x \, dx</math>}}
and collecting the integrals to one side gives
and collecting the integrals to one side gives
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{{Fristående formel||<math>\int e^x \cos x \, dx = {\textstyle\frac{1}{2}}e^x ( \cos x + \sin x) + C\,\mbox{.}</math>}}
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{{Displayed math||<math>\int e^x \cos x \, dx = {\textstyle\frac{1}{2}}e^x ( \cos x + \sin x) + C\,\mbox{.}</math>}}
Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions.
Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions.
Zeile 129: Zeile 129:
The integral can be rewritten as
The integral can be rewritten as
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{{Fristående formel||<math>\int_{0}^{1} \frac{2x}{e^x} \, dx = \int_{0}^{1} 2x \cdot e^{-x} \, dx\,\mbox{.}</math>}}
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{{Displayed math||<math>\int_{0}^{1} \frac{2x}{e^x} \, dx = \int_{0}^{1} 2x \cdot e^{-x} \, dx\,\mbox{.}</math>}}
Substitute <math>f=e^{-x}</math> and <math>g=2x</math>, and integrate by parts
Substitute <math>f=e^{-x}</math> and <math>g=2x</math>, and integrate by parts
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{{Fristående formel||<math>\begin{align*}\int_{0}^{1} 2x \cdot e^{-x} \, dx &= \Bigl[\,-2x\,e^{-x}\,\Bigr]_{0}^{1} + \int_{0}^{1} 2 e^{-x}\,dx\\[4pt] &= \Bigl[\,-2x e^{-x}\,\Bigr]_{0}^{1} + \Bigl[\,-2 e^{-x}\, \Bigr]_{0}^{1}\\[4pt] &= (-2 \cdot e^{-1}) - 0 + (- 2\cdot e^{-1}) - (-2)\\[4pt] &= - \frac{2}{e} - \frac{2}{e} + 2 = 2 - \frac{4}{e}\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*}\int_{0}^{1} 2x \cdot e^{-x} \, dx &= \Bigl[\,-2x\,e^{-x}\,\Bigr]_{0}^{1} + \int_{0}^{1} 2 e^{-x}\,dx\\[4pt] &= \Bigl[\,-2x e^{-x}\,\Bigr]_{0}^{1} + \Bigl[\,-2 e^{-x}\, \Bigr]_{0}^{1}\\[4pt] &= (-2 \cdot e^{-1}) - 0 + (- 2\cdot e^{-1}) - (-2)\\[4pt] &= - \frac{2}{e} - \frac{2}{e} + 2 = 2 - \frac{4}{e}\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Zeile 145: Zeile 145:
We start by performing a variable substitution <math>u=\sqrt{x}</math> which gives <math>du=dx/2\sqrt{x} = dx/2u</math>, that is, <math>dx = 2u\,du\,</math>,
We start by performing a variable substitution <math>u=\sqrt{x}</math> which gives <math>du=dx/2\sqrt{x} = dx/2u</math>, that is, <math>dx = 2u\,du\,</math>,
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{{Fristående formel||<math>\int \ln \sqrt{x} \, dx = \int \ln u \cdot 2u \, du\,\mbox{.}</math>}}
+
{{Displayed math||<math>\int \ln \sqrt{x} \, dx = \int \ln u \cdot 2u \, du\,\mbox{.}</math>}}
Then we integrate by parts. Put <math>f=2u</math> and <math>g=\ln u</math>, which gives
Then we integrate by parts. Put <math>f=2u</math> and <math>g=\ln u</math>, which gives
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{{Fristående formel||<math>\begin{align*}\int \ln u \cdot 2u \, du &= u^2 \ln u - \int u^2 \cdot \frac{1}{u} \, du = u^2 \ln u - \int u\, du\\[4pt] &= u^2 \ln u - \frac{u^2}{2} + C = x \ln \sqrt{x} - \frac {x}{2} + C\\[4pt] &= x \bigl( \ln \sqrt{x} - \tfrac{1}{2} \bigr) + C\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*}\int \ln u \cdot 2u \, du &= u^2 \ln u - \int u^2 \cdot \frac{1}{u} \, du = u^2 \ln u - \int u\, du\\[4pt] &= u^2 \ln u - \frac{u^2}{2} + C = x \ln \sqrt{x} - \frac {x}{2} + C\\[4pt] &= x \bigl( \ln \sqrt{x} - \tfrac{1}{2} \bigr) + C\,\mbox{.}\end{align*}</math>}}
''Note.'' An alternative approach is to rewrite the initial integrand as <math>\ln\sqrt{x} = \tfrac{1}{2}\ln x</math> and then integrate by parts the product <math>\tfrac{1}{2}\cdot\ln x</math>.
''Note.'' An alternative approach is to rewrite the initial integrand as <math>\ln\sqrt{x} = \tfrac{1}{2}\ln x</math> and then integrate by parts the product <math>\tfrac{1}{2}\cdot\ln x</math>.
</div>
</div>

Version vom 08:25, 17. Sep. 2008

       Theory          Exercises      

Contents:

  • Integration by parts.

Learning outcomes:

After this section, you will have learned to:

  • Understand the derivation of the formula for integration by parts.
  • Solve problems about integration that require integration by parts, followed by a substitution (or vice versa).

Integration by parts

To integrate products, one sometimes can make use of a method known as integration by parts. The method is based on the reverse use of the rules for differentiation of products. If f and g are two differentiable functions then the rule for products gives

D(fg)=fg+fg.

Now if one integrates both sides one gets

fg=(fg+fg)dx=fgdx+fgdx 

or after re-ordering

fgdx=fg−fgdx. 

This gives us the formula for integration by parts.

Integration by parts:

f(x)g(x)dx=F(x)g(x)−F(x)g(x)dx. 

This means in practice that one integrates a product of functions by calling one factor f and the other g, and then replaces the integral fgdx  , hopefully, by an easier integral Fgdx,   where F is a primitive function of f and g is the derivative of g.


It is important to note that the method does not always lead to an integral which is easier than the original. It may also be crucial how one chooses the functions f and g, as the following example shows.

Example 1

Determine the integral xsinxdx .

If one chooses f=x and g=sinx one gets F=x22 and g=cosx, and the formula for integration by parts gives

xsinxdx=2x2sinx−2x2cosxdx. 

The new integral on the right-hand side in this case is not easier than the original integral.

If, instead, one chooses f=sinx and g=x then F=−cosx and g=1, and

xsinxdx=−xcosx−−1cosxdx=−xcosx+sinx+C. 

Example 2

Determine the integral  x2lnxdx .

Put f=x2 and g=lnx since differentiation eliminates the logarithm when we carry out an integration by parts: F=x33 and g=1x. This gives us that

x2lnxdx=3x3lnx−3x3x1dx=3x3lnx−31x2dx=3x3lnx−313x3+C=31x3(lnx−31)+C.

Example 3

Determine the integral  x2exdx .

Put f=ex and g=x2, which gives that F=ex and g=2x, and an integration by parts gives

x2exdx=x2ex−2xexdx. 

This requires further integration by parts to solve the new integral 2xexdx . We choose in this case f=ex and g=2x, which gives F=ex and g=2

2xexdx=2xex−2exdx=2xex−2ex+C. 

The original integral thus becomes

x2exdx=x2ex−2xex+2ex+C. 

Example 4

Determine the integral  excosxdx .

In a first integration by parts, we have chosen to integrate the factor ex and differentiate the factor \displaystyle \cos x,

\displaystyle \begin{align*}\int e^x \cos x \, dx &= e^x \cdot \cos x - \int e^x \cdot(-\sin x) \, dx\\[10pt] &= e^x \cos x + \int e^x \sin x \, dx\,\mbox{.}\end{align*}

The result of this is that we essentially have replaced the factor \displaystyle \cos x by \displaystyle \sin x in the integral. If we therefore use integration by parts once again (integrate the \displaystyle e^x and differentiate the \displaystyle \sin x) we get

\displaystyle \int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx\,\mbox{.}

Thus the original integral appears here again. Summarising we have:

\displaystyle \int e^x \cos x \, dx = e^x \cos x + e^x \sin x - \int e^x \cos x \, dx

and collecting the integrals to one side gives

\displaystyle \int e^x \cos x \, dx = {\textstyle\frac{1}{2}}e^x ( \cos x + \sin x) + C\,\mbox{.}

Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions.

Example 5

Determine the integral \displaystyle \ \int_{0}^{1} \frac{2x}{e^x} \, dx\,.

The integral can be rewritten as

\displaystyle \int_{0}^{1} \frac{2x}{e^x} \, dx = \int_{0}^{1} 2x \cdot e^{-x} \, dx\,\mbox{.}

Substitute \displaystyle f=e^{-x} and \displaystyle g=2x, and integrate by parts

\displaystyle \begin{align*}\int_{0}^{1} 2x \cdot e^{-x} \, dx &= \Bigl[\,-2x\,e^{-x}\,\Bigr]_{0}^{1} + \int_{0}^{1} 2 e^{-x}\,dx\\[4pt] &= \Bigl[\,-2x e^{-x}\,\Bigr]_{0}^{1} + \Bigl[\,-2 e^{-x}\, \Bigr]_{0}^{1}\\[4pt] &= (-2 \cdot e^{-1}) - 0 + (- 2\cdot e^{-1}) - (-2)\\[4pt] &= - \frac{2}{e} - \frac{2}{e} + 2 = 2 - \frac{4}{e}\,\mbox{.}\end{align*}

Example 6

Determine the integral \displaystyle \ \int \ln \sqrt{x} \ dx\,.

We start by performing a variable substitution \displaystyle u=\sqrt{x} which gives \displaystyle du=dx/2\sqrt{x} = dx/2u, that is, \displaystyle dx = 2u\,du\,,

\displaystyle \int \ln \sqrt{x} \, dx = \int \ln u \cdot 2u \, du\,\mbox{.}

Then we integrate by parts. Put \displaystyle f=2u and \displaystyle g=\ln u, which gives

\displaystyle \begin{align*}\int \ln u \cdot 2u \, du &= u^2 \ln u - \int u^2 \cdot \frac{1}{u} \, du = u^2 \ln u - \int u\, du\\[4pt] &= u^2 \ln u - \frac{u^2}{2} + C = x \ln \sqrt{x} - \frac {x}{2} + C\\[4pt] &= x \bigl( \ln \sqrt{x} - \tfrac{1}{2} \bigr) + C\,\mbox{.}\end{align*}

Note. An alternative approach is to rewrite the initial integrand as \displaystyle \ln\sqrt{x} = \tfrac{1}{2}\ln x and then integrate by parts the product \displaystyle \tfrac{1}{2}\cdot\ln x.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/2.3_Partielle_Integration“